3.282 \(\int \frac{a+b x^2+c x^4}{(d+e x^2)^{7/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac{x^5 \left (2 e (4 a e+b d)+3 c d^2\right )}{15 d^3 \left (d+e x^2\right )^{5/2}}+\frac{x^3 (4 a e+b d)}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac{a x}{d \left (d+e x^2\right )^{5/2}} \]

[Out]

(a*x)/(d*(d + e*x^2)^(5/2)) + ((b*d + 4*a*e)*x^3)/(3*d^2*(d + e*x^2)^(5/2)) + ((3*c*d^2 + 2*e*(b*d + 4*a*e))*x
^5)/(15*d^3*(d + e*x^2)^(5/2))

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Rubi [A]  time = 0.107052, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1155, 1803, 12, 264} \[ \frac{x^5 \left (2 e (4 a e+b d)+3 c d^2\right )}{15 d^3 \left (d+e x^2\right )^{5/2}}+\frac{x^3 (4 a e+b d)}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac{a x}{d \left (d+e x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^(7/2),x]

[Out]

(a*x)/(d*(d + e*x^2)^(5/2)) + ((b*d + 4*a*e)*x^3)/(3*d^2*(d + e*x^2)^(5/2)) + ((3*c*d^2 + 2*e*(b*d + 4*a*e))*x
^5)/(15*d^3*(d + e*x^2)^(5/2))

Rule 1155

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(a^p*x*(d + e*x^2
)^(q + 1))/d, x] + Dist[1/d, Int[x^2*(d + e*x^2)^q*(d*PolynomialQuotient[(a + b*x^2 + c*x^4)^p - a^p, x^2, x]
- e*a^p*(2*q + 3)), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0
] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4*p + 2*q + 1, 0]

Rule 1803

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{A = Coeff[Pq, x, 0], Q = PolynomialQuotient
[Pq - Coeff[Pq, x, 0], x^2, x]}, Simp[(A*x^(m + 1)*(a + b*x^2)^(p + 1))/(a*(m + 1)), x] + Dist[1/(a*(m + 1)),
Int[x^(m + 2)*(a + b*x^2)^p*(a*(m + 1)*Q - A*b*(m + 2*(p + 1) + 1)), x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x^2] && IntegerQ[m/2] && ILtQ[(m + 1)/2 + p, 0] && LtQ[m + Expon[Pq, x] + 2*p + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx &=\frac{a x}{d \left (d+e x^2\right )^{5/2}}+\frac{\int \frac{x^2 \left (4 a e+d \left (b+c x^2\right )\right )}{\left (d+e x^2\right )^{7/2}} \, dx}{d}\\ &=\frac{a x}{d \left (d+e x^2\right )^{5/2}}+\frac{(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac{\int \frac{\left (3 c d^2+2 e (b d+4 a e)\right ) x^4}{\left (d+e x^2\right )^{7/2}} \, dx}{3 d^2}\\ &=\frac{a x}{d \left (d+e x^2\right )^{5/2}}+\frac{(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac{1}{3} \left (3 c+\frac{2 e (b d+4 a e)}{d^2}\right ) \int \frac{x^4}{\left (d+e x^2\right )^{7/2}} \, dx\\ &=\frac{a x}{d \left (d+e x^2\right )^{5/2}}+\frac{(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac{\left (3 c d^2+2 e (b d+4 a e)\right ) x^5}{15 d^3 \left (d+e x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.051325, size = 67, normalized size = 0.78 \[ \frac{a \left (15 d^2 x+20 d e x^3+8 e^2 x^5\right )+d x^3 \left (5 b d+2 b e x^2+3 c d x^2\right )}{15 d^3 \left (d+e x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^(7/2),x]

[Out]

(d*x^3*(5*b*d + 3*c*d*x^2 + 2*b*e*x^2) + a*(15*d^2*x + 20*d*e*x^3 + 8*e^2*x^5))/(15*d^3*(d + e*x^2)^(5/2))

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Maple [A]  time = 0.006, size = 66, normalized size = 0.8 \begin{align*}{\frac{x \left ( 8\,a{e}^{2}{x}^{4}+2\,bde{x}^{4}+3\,c{d}^{2}{x}^{4}+20\,ade{x}^{2}+5\,b{d}^{2}{x}^{2}+15\,a{d}^{2} \right ) }{15\,{d}^{3}} \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/(e*x^2+d)^(7/2),x)

[Out]

1/15*x*(8*a*e^2*x^4+2*b*d*e*x^4+3*c*d^2*x^4+20*a*d*e*x^2+5*b*d^2*x^2+15*a*d^2)/(e*x^2+d)^(5/2)/d^3

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Maxima [B]  time = 0.952352, size = 234, normalized size = 2.72 \begin{align*} -\frac{c x^{3}}{2 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} e} + \frac{8 \, a x}{15 \, \sqrt{e x^{2} + d} d^{3}} + \frac{4 \, a x}{15 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{2}} + \frac{a x}{5 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} d} + \frac{c x}{10 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} e^{2}} + \frac{c x}{5 \, \sqrt{e x^{2} + d} d e^{2}} - \frac{3 \, c d x}{10 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} e^{2}} - \frac{b x}{5 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} e} + \frac{2 \, b x}{15 \, \sqrt{e x^{2} + d} d^{2} e} + \frac{b x}{15 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(7/2),x, algorithm="maxima")

[Out]

-1/2*c*x^3/((e*x^2 + d)^(5/2)*e) + 8/15*a*x/(sqrt(e*x^2 + d)*d^3) + 4/15*a*x/((e*x^2 + d)^(3/2)*d^2) + 1/5*a*x
/((e*x^2 + d)^(5/2)*d) + 1/10*c*x/((e*x^2 + d)^(3/2)*e^2) + 1/5*c*x/(sqrt(e*x^2 + d)*d*e^2) - 3/10*c*d*x/((e*x
^2 + d)^(5/2)*e^2) - 1/5*b*x/((e*x^2 + d)^(5/2)*e) + 2/15*b*x/(sqrt(e*x^2 + d)*d^2*e) + 1/15*b*x/((e*x^2 + d)^
(3/2)*d*e)

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Fricas [A]  time = 4.71996, size = 198, normalized size = 2.3 \begin{align*} \frac{{\left ({\left (3 \, c d^{2} + 2 \, b d e + 8 \, a e^{2}\right )} x^{5} + 15 \, a d^{2} x + 5 \,{\left (b d^{2} + 4 \, a d e\right )} x^{3}\right )} \sqrt{e x^{2} + d}}{15 \,{\left (d^{3} e^{3} x^{6} + 3 \, d^{4} e^{2} x^{4} + 3 \, d^{5} e x^{2} + d^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(7/2),x, algorithm="fricas")

[Out]

1/15*((3*c*d^2 + 2*b*d*e + 8*a*e^2)*x^5 + 15*a*d^2*x + 5*(b*d^2 + 4*a*d*e)*x^3)*sqrt(e*x^2 + d)/(d^3*e^3*x^6 +
3*d^4*e^2*x^4 + 3*d^5*e*x^2 + d^6)

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Sympy [B]  time = 56.603, size = 639, normalized size = 7.43 \begin{align*} a \left (\frac{15 d^{5} x}{15 d^{\frac{17}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{15}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{13}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{11}{2}} e^{3} x^{6} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{35 d^{4} e x^{3}}{15 d^{\frac{17}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{15}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{13}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{11}{2}} e^{3} x^{6} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{28 d^{3} e^{2} x^{5}}{15 d^{\frac{17}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{15}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{13}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{11}{2}} e^{3} x^{6} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{8 d^{2} e^{3} x^{7}}{15 d^{\frac{17}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{15}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 45 d^{\frac{13}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{11}{2}} e^{3} x^{6} \sqrt{1 + \frac{e x^{2}}{d}}}\right ) + b \left (\frac{5 d x^{3}}{15 d^{\frac{9}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 30 d^{\frac{7}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{5}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{2 e x^{5}}{15 d^{\frac{9}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 30 d^{\frac{7}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 15 d^{\frac{5}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}}}\right ) + \frac{c x^{5}}{5 d^{\frac{7}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 10 d^{\frac{5}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}} + 5 d^{\frac{3}{2}} e^{2} x^{4} \sqrt{1 + \frac{e x^{2}}{d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/(e*x**2+d)**(7/2),x)

[Out]

a*(15*d**5*x/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**
4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d)) + 35*d**4*e*x**3/(15*d**(17/2)*sqrt(1 + e*x*
*2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3
*x**6*sqrt(1 + e*x**2/d)) + 28*d**3*e**2*x**5/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 +
e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d)) + 8*d**2*e*
*3*x**7/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqr
t(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d))) + b*(5*d*x**3/(15*d**(9/2)*sqrt(1 + e*x**2/d) +
30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d**(5/2)*e**2*x**4*sqrt(1 + e*x**2/d)) + 2*e*x**5/(15*d**(9/2)*sqrt
(1 + e*x**2/d) + 30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d**(5/2)*e**2*x**4*sqrt(1 + e*x**2/d))) + c*x**5/(
5*d**(7/2)*sqrt(1 + e*x**2/d) + 10*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d) + 5*d**(3/2)*e**2*x**4*sqrt(1 + e*x**2/d
))

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Giac [A]  time = 1.16173, size = 101, normalized size = 1.17 \begin{align*} \frac{{\left (x^{2}{\left (\frac{{\left (3 \, c d^{2} e^{2} + 2 \, b d e^{3} + 8 \, a e^{4}\right )} x^{2} e^{\left (-2\right )}}{d^{3}} + \frac{5 \,{\left (b d^{2} e^{2} + 4 \, a d e^{3}\right )} e^{\left (-2\right )}}{d^{3}}\right )} + \frac{15 \, a}{d}\right )} x}{15 \,{\left (x^{2} e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(7/2),x, algorithm="giac")

[Out]

1/15*(x^2*((3*c*d^2*e^2 + 2*b*d*e^3 + 8*a*e^4)*x^2*e^(-2)/d^3 + 5*(b*d^2*e^2 + 4*a*d*e^3)*e^(-2)/d^3) + 15*a/d
)*x/(x^2*e + d)^(5/2)